Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ .

Given three positive numbers $$a,\,b,\,c$$ . Prove that $$(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$$ .

My own problem is given a solution, and I'm looking forward to seeing a nicer one(s), thank you !

Solution. Without loss of generality, we may suppose $$(a- 1)(b- 1)\geqq 0 \therefore 1+ ab\geqq a+ b$$. Then

$$(abc+ a+ b+ c)^{3}= \left ( c(1+ ab)+ a+ b \right )^{2}\left ( c(1+ ab)+ a+ b \right )\geqq$$

$$\geqq 4\,c(1+ ab)(a+ b)\left ( c(a+ b)+ (a+ b) \right )= 2\,c(a+ b)^{2}(2+ 2\,ab)(1+ c)$$

Again, by using a.m.-g.m.-inequality, we have $$2c(\!a+ b\!)^{2}(\!2+ 2ab\!)(\!1+ c\!)\geqq 8abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$$

q.e.d

2 Answers

After replacing $$a\rightarrow\frac{1}{a},$$ $$b\rightarrow\frac{1}{b}$$ and $$c\rightarrow\frac{1}{c}$$ we need to prove that $$(1+ab+ac+bc)^3\geq8abc(1+a)(1+b)(1+c).$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2,$$ where $$v>0$$ and $$abc=w^3$$.

Thus, we need to prove that: $$(1+3v^2)^3\geq8w^3(1+w^3+3v^2+3u)$$ or $$(1+3v^2)\geq8(w^3+w^6+3v^3w^2+3uw^3)$$ and since by AM-GM $$v\geq w$$ and $$v^4\geq uw^3,$$ it's enough to prove that: $$(1+3v^2)^3\geq8(v^3+v^6+3v^5+3v^4)$$ or $$(1+3v^2)^3\geq8v^3(1+v)^3$$ or $$1+3v^2\geq2v(1+v)$$ or $$(1-v)^2\geq0$$ and we are done!

Also, $$uvw$$ kills this inequality, but it's not so nice.

• YOU'RE SO SUPER ! What makes you think to replace by a substitution? – user680032 Jun 23 at 5:15
• I think your solution is the neatest and the best one ! – user680032 Jun 23 at 5:15
• @HanhVyheaven28 Because after this substitution the inequality looks simpler.Thank you! – Michael Rozenberg Jun 23 at 5:16
• Wow, that's so impressive! I found it by discriminant and you do this easily! – user680032 Jun 23 at 5:18

Hint: By AM-GM, $$(abc+a) + (b+c) \geqslant 2a\sqrt{bc}+2\sqrt{bc} = 2\sqrt{bc}(1+a)$$

Now multiply three such inequalities (after cyclical shift).

• Very nice! Bravo! +1 – Michael Rozenberg Jun 23 at 7:38
• 'Now multiply three such inequalities (after cyclical shift).', that made a big impres ! – user680032 Jun 23 at 9:12